ED-9123 Mechanisms Design and Simulation DR.V.SUNDARESWARAN PROFESSOR OF MECHANICAL ENGG. COLLEGE OF ENGINEERING, GUINDY, ANNA UNIVERSITY, CHENNAI – 600 025 4.1.2012

GROSS MOTION ANALYSIS OF MECHANISMS Given a mechanism, to find out the motion capabilities of the input and the output links. In a four bar mechanism, this is simple in the case of Double Crank mechanism. As both the links have full rotation, analysis is not needed. In Crank Rocker mechanism, input has full rotation. To find out the rocker motion capability, ie., swing / oscillation angle, analysis has to be done. Limit positions of the output / rocker has to be obtained.

Right Extreme position of the output is one limit position Right Extreme position of the output is one limit position. When the input link and the coupler are extended as a straight line, this occurs. Left Extreme position of the output is another limit position. This position is obtained, when the input link and coupler are folded as a straight line. In the Double Rocker mechanisms, as there are two rockers, two swing angles have to be determined. First, assume one of the rocker as input and do the procedure mentioned above, which will give the swing angle of its output link. Then, make this as the input link and repeat the procedure to get the swing angle of the second rocker.

Analysis of the above configuration, can be used to design a four bar crank rocker quick return mechanism. In a quick return mechanism, the forward stroke of the output takes longer time than the return stroke. Ratio of these times is known as time ratio. Time taken for forward/cutting stroke Time Ratio (T.R) = ------------------------------------------------ Time taken for return/idle stroke Time Ratio > 1 - Quick Return Mechanism Time Ratio = 1 - Unit Time-Ratio Mechanism

Time Ratio = 1.25 Swing Angle,Φ = 75⁰ Design a four bar crank rocker quick return mechanism for the following data: Time Ratio = 1.25 Swing Angle,Φ = 75⁰ Note: These two are the necessary data for this problem. Additionally, output link length, “c” may, also, be given. PROCEDURE: 1. Calculate the angle θ, using the expression 180 + θ Time Ratio (T.R)= -------------- 180 – θ For T.R = 1.25, θ = 20⁰

2. Assume an arbitrary location for output link pivot, B₀ , and locate it. 3. Using the given / assumed output link length (say 60 mm), draw the output link , B₀B, arbitrarily in its right extreme position, B₀B1. 4. Then, draw the output link, in its left extreme position, B₀B2, using the given swing angle,Φ. In this case 75⁰ 5. Draw an arbitrary line B1X , such that X is left of output link pivot. 6. Measure angle, γ , at the intersection of this arbitrary line with the left extreme position line B₀B2.

7. Draw the line B2Z, such that the intersection of this line with the line B1X an angle θ. This can be done, by knowing the angle δ, which is equal to ( γ – θ ). Name the intersecting point of the above two lines as A0, ie., the pivot of the input crank. 8. Measure the length, A0B1, which is equal to the sum of input link length and coupler length, ie., ( a + b). 9. Measure the length, A0B2, which is equal to the difference of coupler and input link lengths, ie.,( b – a ). Then, A0B1 - A0B2 = ( a + b) - ( b – a ) = 2 a. Half of this length is the input link length, a. Then, find the coupler length, b, from the expression, b = (A0B1 – a)

10. Join the points A0 and B0. Measure the length A0B0, which is the fixed link length, d. 11. Sketch the designed four-bar mechanism with arbitrary input link position, and indicate the determined dimensions. 12. As there are two arbitrary lines, B0B1 and B1X there will be infinite number of solutions for this problem. Better solutions / Best solution can be obtained considering the following: 1. Link length Ratio and 2. Transmission angle. 13. Link length ratio = ( Lmax / Lmin ) where, Lmax is the longest link length and Lmin is the shortest link length. 14. Transmission angle, μ , should be closer to 90⁰.

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